You want to make five-letter codes that use the letters a, f, e, r, and m without repeating any letter. what is the probability that a randomly chosen code starts with m and ends with an e

Given: Five letter codes are to be made with the letter A, F, E, R and M without repitition.

To Find: Probability that a randomly chosen code starts with M and ends with E.

Solution: So, required probability is .

Calculation/Explanation:

Let us calculate the required probability step-wise.

To pick the first letter as M, there are 5 possible outcomes (A, F, E, R or M) and picking M is only one possibility out 5 outcomes.

So, probability of picking an M first is equal to ...(1)

After this, in the second, third and fourth places, we must not pick an E. In the second place, there are 4 possible outcomes (A, F, E or R) out of which not picking an E means there are 3 possibilities (A, F or R) out of the 4 outcomes. So, for the second letter the probability of not picking an E is ...(2)

For the third letter, there are now 3 possible outcomes (since out 5 letters, 2 have already been chosen). One of these outcomes is an E and two will be non-E. This means, probability of not drawing an E is ...(3)

For the fourth letter, there are now 2 possible outcomes out of which one is an E and one is not an E. This means, the probability of not picking an E is ...(4)

For the final letter, only one possible outcome remains which is E. And therefore the probability that we draw an E would be 1 ...(5)

To get the total probability of starting the code with M and ending with E, we multiply the individual probabilities i.e., (1), (2), (3), (4) and (5).

So, required probability is .

answer is x<

the answer is (-4,-4)

hope this ! : d